Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long> > matr(n, vector<long long>(m, 0));
  for (auto &x : matr)
    for (auto &y : x) cin >> y;
  long long answ = 0;
  vector<pair<long long, long long> > ra, ca;
  if (m > n) {
    for (long long i = 0; i < n; i++) {
      long long mn = 1e18 + 7;
      for (long long j = 0; j < m; j++) {
        mn = min(mn, matr[i][j]);
      }
      if (mn > 0) ra.push_back({i + 1, mn});
      answ += mn;
      for (long long j = 0; j < m; j++) {
        matr[i][j] -= mn;
      }
    }
    for (long long j = 0; j < m; j++) {
      long long mn = 1e18 + 7;
      for (long long i = 0; i < n; i++) {
        mn = min(mn, matr[i][j]);
      }
      if (mn > 0) ca.push_back({j + 1, mn});
      answ += mn;
      for (long long i = 0; i < n; i++) {
        matr[i][j] -= mn;
      }
    }
  } else {
    for (long long j = 0; j < m; j++) {
      long long mn = 1e18 + 7;
      for (long long i = 0; i < n; i++) {
        mn = min(mn, matr[i][j]);
      }
      if (mn > 0) ca.push_back({j + 1, mn});
      answ += mn;
      for (long long i = 0; i < n; i++) {
        matr[i][j] -= mn;
      }
    }
    for (long long i = 0; i < n; i++) {
      long long mn = 1e18 + 7;
      for (long long j = 0; j < m; j++) {
        mn = min(mn, matr[i][j]);
      }
      if (mn > 0) ra.push_back({i + 1, mn});
      answ += mn;
      for (long long j = 0; j < m; j++) {
        matr[i][j] -= mn;
      }
    }
  }
  bool flag = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++)
      if (matr[i][j]) flag = 1;
  if (flag)
    cout << -1;
  else {
    cout << answ << endl;
    for (auto &x : ra) {
      for (long long i = 0; i < x.second; i++)
        cout << "row " << x.first << endl;
    }
    for (auto &x : ca) {
      for (long long i = 0; i < x.second; i++)
        cout << "col " << x.first << endl;
    }
  }
  cin >> n;
  return 0;
}